| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Geometric curve properties |
| Difficulty | Standard +0.8 This question requires geometric interpretation to set up the differential equation (relating triangle area to coordinates and gradient), then solving a separable DE with trigonometric functions. The geometric setup in part (i) demands careful spatial reasoning beyond routine calculus, and part (ii) involves non-trivial integration of cosec²x. More challenging than standard DE questions but accessible to strong P3 students. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(\frac{y}{TN} = \frac{dy}{dx}\), or equivalent | B1 | |
| Express area of \(PTN\) in terms of \(y\) and \(\frac{dy}{dx}\), and equate to \(\tan x\) | M1 | |
| Obtain given relation correctly | A1 | [3 marks] |
| (ii) Separate variables correctly | B1 | |
| Integrate and obtain term \(-\frac{2}{y}\), or equivalent | B1 | |
| Integrate and obtain term \(\ln(\sin x)\), or equivalent | B1 | |
| Evaluate a constant or use limits \(y = 2\), \(x = \frac{\pi}{6}\) in a solution containing a term of the form \(a/y\) or \(b\ln(\sin x)\) | M1 | |
| Obtain correct solution in any form, e.g. \(-\frac{2}{y} = \ln(2\sin x) - 1\) | A1 | |
| Rearrange as \(y = 2/(1 - \ln(2\sin x))\), or equivalent | A1 | [Allow decimals, e.g. as in a solution \(y = 2/(0.3 - \ln(\sin x))\).] |
**(i)** State $\frac{y}{TN} = \frac{dy}{dx}$, or equivalent | B1 |
Express area of $PTN$ in terms of $y$ and $\frac{dy}{dx}$, and equate to $\tan x$ | M1 |
Obtain given relation correctly | A1 | [3 marks]
**(ii)** Separate variables correctly | B1 |
Integrate and obtain term $-\frac{2}{y}$, or equivalent | B1 |
Integrate and obtain term $\ln(\sin x)$, or equivalent | B1 |
Evaluate a constant or use limits $y = 2$, $x = \frac{\pi}{6}$ in a solution containing a term of the form $a/y$ or $b\ln(\sin x)$ | M1 |
Obtain correct solution in any form, e.g. $-\frac{2}{y} = \ln(2\sin x) - 1$ | A1 |
Rearrange as $y = 2/(1 - \ln(2\sin x))$, or equivalent | A1 | [Allow decimals, e.g. as in a solution $y = 2/(0.3 - \ln(\sin x))$.]
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\includegraphics[max width=\textwidth, alt={}, center]{20893bfc-3300-4205-9d2c-729cc3243971-3_597_951_1471_598}
In the diagram the tangent to a curve at a general point $P$ with coordinates $( x , y )$ meets the $x$-axis at $T$. The point $N$ on the $x$-axis is such that $P N$ is perpendicular to the $x$-axis. The curve is such that, for all values of $x$ in the interval $0 < x < \frac { 1 } { 2 } \pi$, the area of triangle $P T N$ is equal to $\tan x$, where $x$ is in radians.\\
(i) Using the fact that the gradient of the curve at $P$ is $\frac { P N } { T N }$, show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 } y ^ { 2 } \cot x .$$
(ii) Given that $y = 2$ when $x = \frac { 1 } { 6 } \pi$, solve this differential equation to find the equation of the curve, expressing $y$ in terms of $x$.
\hfill \mbox{\textit{CAIE P3 2008 Q8 [9]}}